274288
Consider that a $\mathrm{d}^{6}$ metal ion $\left[\mathrm{M}^{2+}\right]$ forms a complex with aqua ligands and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilisation energy of the complex is
1 tetrahedral and $-1.6 \Delta_{\mathrm{t}}+1 \mathrm{P}$
2 octahedral and $-2.4 \Delta_{0}+2 \mathrm{P}$
3 octahedral and $-1.6 \Delta_{0}$
4 tetrahedral and $-0.6 \Delta_{\mathrm{t}}$
Explanation:
(D) : The metal ion $\left(\mathrm{M}^{2+}\right)$ forms the complex with aqua ligands that means it contain the weak field ligand. It is a iron metal which has the following electronic configuration- $\mathrm{Fe}^{2 \mathrm{t}}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ If metal forms the tetrahedral complex then it contains the four unpaired electron. \(\text { CFSE }=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.4 \times \mathrm{N}_{\mathrm{t}_2}\right] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-0.6 \times 3+0.4 \times 3] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-1.8+1.2] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=-0.6 \Delta_{\mathrm{t}}\) Hence, the geometry and the crystal field stabilisation energy of the complex is tetrahedral and \(-0.6 \Delta_t\) respectively.
JEE Main 2020
COORDINATION COMPOUNDS
274289
The electronic spectrum of $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ shows a single broad peak with a maximum at 20,300 cm-1. The crystal field stabilization energy (CFSE) of the complex ion, in $\mathrm{kJ} \mathrm{mol}^{-1}$, is
1 145.5
2 242.5
3 83.7
4 97
Explanation:
(D) : Given: } \Delta_{\mathrm{o}}=20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=\text { ? }$ $\because \quad 1 \mathrm{Kj} / \mathrm{mol}=83.7 \mathrm{~cm}^{-1}$ ${\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Ti}$. $\therefore \quad x+6(0)=+3$ $x=+3$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{eg}_{\mathrm{g}}}\right] \Delta_{\mathrm{o}}$ $\mathrm{CFSE}=[-0.4 \times 1+0.6 \times 0] \Delta_{\mathrm{o}}$ Or $\mathrm{CFSE}=0.4 \Delta_{\mathrm{o}}$ (ignoring the negtive sign $)$ CFSE $=0.4 \times 20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=0.4 \times 20,300 \times \frac{1}{83.7} \mathrm{~kJ} / \mathrm{mol}$ $\mathrm{CFSE}=97 \mathrm{~kJ} \mathrm{~mol}^{-1}$
JEE Main 2020
COORDINATION COMPOUNDS
274290
The d-electron configuration of $\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$, respectively are
1 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
2 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
3 $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
4 $t_{2 g}^{4} e_{g}^{2}$ and $t_{2 g}^{4} e_{g}^{2}$
Explanation:
(B) : {$\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$} Let, $x$ be the oxidation state of Ru. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $\mathrm{x}$ be the oxidation state of Fe. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\because$ Pairing occure because 'Ru' belongs to the $4 \mathrm{~d} \quad \mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ (High spin) -series metal. $\mathrm{Ru}^{2+}[\mathrm{Kr}] 4 \mathrm{~d}^{6} 55^{\circ}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
JEE Main 2020
COORDINATION COMPOUNDS
274295
The complex that has highest crystal field splitting energy $(\Delta)$, is
(C) : It is clear that the CFSE of tetrahedral complex are always less than the octahedral complex. Thus, option (d) has less CFSE than option (a), (b) and (c) because it is a tetrahedral complex. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ forms the low spin complex because it contains the SFL. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ $3(+1)+\mathrm{x}+6(-1)=0$ $\mathrm{x}=+3$ $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ $\mathrm{d}^{6}-\mathrm{C}_{\mathrm{g}}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 g}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{\text {o }}$ $\mathrm{CFSE}=[-0.4 \times 6+0.6 \times 0] \Delta_{\text {。 }}$ $\mathrm{CFSE}=-2.4 \Delta_{\mathrm{o}}$ Hence, $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ has highest crystal field splitting energy.
JEE Main 2019
COORDINATION COMPOUNDS
274296
The crystal field stabilization energy (CFSE) of $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$, respectively, are
1 $-0.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
2 $-0.4 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
3 $-2.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
4 $-0.6 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
Explanation:
(B) : $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{NiCl}_{4}\right]$ are octahedral and tetrahedral complex respectively. $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $x$ be the oxidation state of Fe. $x+6(0)+2(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{0}$ CFSE $=[-0.4 \times 4+0.6 \times 2] \Delta_{0}$ $\mathrm{CFSE}=-0.4 \Delta_{0}$ and, for $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$ $2(+1)+x+4(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}$ $\mathrm{d}^{8}-\uparrow_{\uparrow \downarrow} \uparrow \uparrow_{\mathrm{t}_{2}}$ CFSE $=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.6 \times \mathrm{N}_{\mathrm{t}_{2}}\right] \Delta_{\mathrm{t}}$ CFSE $=[-0.6 \times 4+0.6 \times 4] \Delta_{\mathrm{t}}$ $\mathrm{CFSE}=-0.8 \Delta_{\mathrm{t}}$
274288
Consider that a $\mathrm{d}^{6}$ metal ion $\left[\mathrm{M}^{2+}\right]$ forms a complex with aqua ligands and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilisation energy of the complex is
1 tetrahedral and $-1.6 \Delta_{\mathrm{t}}+1 \mathrm{P}$
2 octahedral and $-2.4 \Delta_{0}+2 \mathrm{P}$
3 octahedral and $-1.6 \Delta_{0}$
4 tetrahedral and $-0.6 \Delta_{\mathrm{t}}$
Explanation:
(D) : The metal ion $\left(\mathrm{M}^{2+}\right)$ forms the complex with aqua ligands that means it contain the weak field ligand. It is a iron metal which has the following electronic configuration- $\mathrm{Fe}^{2 \mathrm{t}}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ If metal forms the tetrahedral complex then it contains the four unpaired electron. \(\text { CFSE }=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.4 \times \mathrm{N}_{\mathrm{t}_2}\right] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-0.6 \times 3+0.4 \times 3] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-1.8+1.2] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=-0.6 \Delta_{\mathrm{t}}\) Hence, the geometry and the crystal field stabilisation energy of the complex is tetrahedral and \(-0.6 \Delta_t\) respectively.
JEE Main 2020
COORDINATION COMPOUNDS
274289
The electronic spectrum of $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ shows a single broad peak with a maximum at 20,300 cm-1. The crystal field stabilization energy (CFSE) of the complex ion, in $\mathrm{kJ} \mathrm{mol}^{-1}$, is
1 145.5
2 242.5
3 83.7
4 97
Explanation:
(D) : Given: } \Delta_{\mathrm{o}}=20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=\text { ? }$ $\because \quad 1 \mathrm{Kj} / \mathrm{mol}=83.7 \mathrm{~cm}^{-1}$ ${\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Ti}$. $\therefore \quad x+6(0)=+3$ $x=+3$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{eg}_{\mathrm{g}}}\right] \Delta_{\mathrm{o}}$ $\mathrm{CFSE}=[-0.4 \times 1+0.6 \times 0] \Delta_{\mathrm{o}}$ Or $\mathrm{CFSE}=0.4 \Delta_{\mathrm{o}}$ (ignoring the negtive sign $)$ CFSE $=0.4 \times 20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=0.4 \times 20,300 \times \frac{1}{83.7} \mathrm{~kJ} / \mathrm{mol}$ $\mathrm{CFSE}=97 \mathrm{~kJ} \mathrm{~mol}^{-1}$
JEE Main 2020
COORDINATION COMPOUNDS
274290
The d-electron configuration of $\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$, respectively are
1 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
2 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
3 $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
4 $t_{2 g}^{4} e_{g}^{2}$ and $t_{2 g}^{4} e_{g}^{2}$
Explanation:
(B) : {$\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$} Let, $x$ be the oxidation state of Ru. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $\mathrm{x}$ be the oxidation state of Fe. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\because$ Pairing occure because 'Ru' belongs to the $4 \mathrm{~d} \quad \mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ (High spin) -series metal. $\mathrm{Ru}^{2+}[\mathrm{Kr}] 4 \mathrm{~d}^{6} 55^{\circ}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
JEE Main 2020
COORDINATION COMPOUNDS
274295
The complex that has highest crystal field splitting energy $(\Delta)$, is
(C) : It is clear that the CFSE of tetrahedral complex are always less than the octahedral complex. Thus, option (d) has less CFSE than option (a), (b) and (c) because it is a tetrahedral complex. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ forms the low spin complex because it contains the SFL. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ $3(+1)+\mathrm{x}+6(-1)=0$ $\mathrm{x}=+3$ $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ $\mathrm{d}^{6}-\mathrm{C}_{\mathrm{g}}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 g}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{\text {o }}$ $\mathrm{CFSE}=[-0.4 \times 6+0.6 \times 0] \Delta_{\text {。 }}$ $\mathrm{CFSE}=-2.4 \Delta_{\mathrm{o}}$ Hence, $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ has highest crystal field splitting energy.
JEE Main 2019
COORDINATION COMPOUNDS
274296
The crystal field stabilization energy (CFSE) of $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$, respectively, are
1 $-0.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
2 $-0.4 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
3 $-2.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
4 $-0.6 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
Explanation:
(B) : $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{NiCl}_{4}\right]$ are octahedral and tetrahedral complex respectively. $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $x$ be the oxidation state of Fe. $x+6(0)+2(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{0}$ CFSE $=[-0.4 \times 4+0.6 \times 2] \Delta_{0}$ $\mathrm{CFSE}=-0.4 \Delta_{0}$ and, for $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$ $2(+1)+x+4(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}$ $\mathrm{d}^{8}-\uparrow_{\uparrow \downarrow} \uparrow \uparrow_{\mathrm{t}_{2}}$ CFSE $=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.6 \times \mathrm{N}_{\mathrm{t}_{2}}\right] \Delta_{\mathrm{t}}$ CFSE $=[-0.6 \times 4+0.6 \times 4] \Delta_{\mathrm{t}}$ $\mathrm{CFSE}=-0.8 \Delta_{\mathrm{t}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
COORDINATION COMPOUNDS
274288
Consider that a $\mathrm{d}^{6}$ metal ion $\left[\mathrm{M}^{2+}\right]$ forms a complex with aqua ligands and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilisation energy of the complex is
1 tetrahedral and $-1.6 \Delta_{\mathrm{t}}+1 \mathrm{P}$
2 octahedral and $-2.4 \Delta_{0}+2 \mathrm{P}$
3 octahedral and $-1.6 \Delta_{0}$
4 tetrahedral and $-0.6 \Delta_{\mathrm{t}}$
Explanation:
(D) : The metal ion $\left(\mathrm{M}^{2+}\right)$ forms the complex with aqua ligands that means it contain the weak field ligand. It is a iron metal which has the following electronic configuration- $\mathrm{Fe}^{2 \mathrm{t}}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ If metal forms the tetrahedral complex then it contains the four unpaired electron. \(\text { CFSE }=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.4 \times \mathrm{N}_{\mathrm{t}_2}\right] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-0.6 \times 3+0.4 \times 3] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-1.8+1.2] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=-0.6 \Delta_{\mathrm{t}}\) Hence, the geometry and the crystal field stabilisation energy of the complex is tetrahedral and \(-0.6 \Delta_t\) respectively.
JEE Main 2020
COORDINATION COMPOUNDS
274289
The electronic spectrum of $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ shows a single broad peak with a maximum at 20,300 cm-1. The crystal field stabilization energy (CFSE) of the complex ion, in $\mathrm{kJ} \mathrm{mol}^{-1}$, is
1 145.5
2 242.5
3 83.7
4 97
Explanation:
(D) : Given: } \Delta_{\mathrm{o}}=20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=\text { ? }$ $\because \quad 1 \mathrm{Kj} / \mathrm{mol}=83.7 \mathrm{~cm}^{-1}$ ${\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Ti}$. $\therefore \quad x+6(0)=+3$ $x=+3$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{eg}_{\mathrm{g}}}\right] \Delta_{\mathrm{o}}$ $\mathrm{CFSE}=[-0.4 \times 1+0.6 \times 0] \Delta_{\mathrm{o}}$ Or $\mathrm{CFSE}=0.4 \Delta_{\mathrm{o}}$ (ignoring the negtive sign $)$ CFSE $=0.4 \times 20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=0.4 \times 20,300 \times \frac{1}{83.7} \mathrm{~kJ} / \mathrm{mol}$ $\mathrm{CFSE}=97 \mathrm{~kJ} \mathrm{~mol}^{-1}$
JEE Main 2020
COORDINATION COMPOUNDS
274290
The d-electron configuration of $\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$, respectively are
1 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
2 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
3 $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
4 $t_{2 g}^{4} e_{g}^{2}$ and $t_{2 g}^{4} e_{g}^{2}$
Explanation:
(B) : {$\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$} Let, $x$ be the oxidation state of Ru. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $\mathrm{x}$ be the oxidation state of Fe. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\because$ Pairing occure because 'Ru' belongs to the $4 \mathrm{~d} \quad \mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ (High spin) -series metal. $\mathrm{Ru}^{2+}[\mathrm{Kr}] 4 \mathrm{~d}^{6} 55^{\circ}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
JEE Main 2020
COORDINATION COMPOUNDS
274295
The complex that has highest crystal field splitting energy $(\Delta)$, is
(C) : It is clear that the CFSE of tetrahedral complex are always less than the octahedral complex. Thus, option (d) has less CFSE than option (a), (b) and (c) because it is a tetrahedral complex. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ forms the low spin complex because it contains the SFL. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ $3(+1)+\mathrm{x}+6(-1)=0$ $\mathrm{x}=+3$ $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ $\mathrm{d}^{6}-\mathrm{C}_{\mathrm{g}}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 g}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{\text {o }}$ $\mathrm{CFSE}=[-0.4 \times 6+0.6 \times 0] \Delta_{\text {。 }}$ $\mathrm{CFSE}=-2.4 \Delta_{\mathrm{o}}$ Hence, $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ has highest crystal field splitting energy.
JEE Main 2019
COORDINATION COMPOUNDS
274296
The crystal field stabilization energy (CFSE) of $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$, respectively, are
1 $-0.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
2 $-0.4 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
3 $-2.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
4 $-0.6 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
Explanation:
(B) : $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{NiCl}_{4}\right]$ are octahedral and tetrahedral complex respectively. $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $x$ be the oxidation state of Fe. $x+6(0)+2(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{0}$ CFSE $=[-0.4 \times 4+0.6 \times 2] \Delta_{0}$ $\mathrm{CFSE}=-0.4 \Delta_{0}$ and, for $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$ $2(+1)+x+4(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}$ $\mathrm{d}^{8}-\uparrow_{\uparrow \downarrow} \uparrow \uparrow_{\mathrm{t}_{2}}$ CFSE $=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.6 \times \mathrm{N}_{\mathrm{t}_{2}}\right] \Delta_{\mathrm{t}}$ CFSE $=[-0.6 \times 4+0.6 \times 4] \Delta_{\mathrm{t}}$ $\mathrm{CFSE}=-0.8 \Delta_{\mathrm{t}}$
274288
Consider that a $\mathrm{d}^{6}$ metal ion $\left[\mathrm{M}^{2+}\right]$ forms a complex with aqua ligands and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilisation energy of the complex is
1 tetrahedral and $-1.6 \Delta_{\mathrm{t}}+1 \mathrm{P}$
2 octahedral and $-2.4 \Delta_{0}+2 \mathrm{P}$
3 octahedral and $-1.6 \Delta_{0}$
4 tetrahedral and $-0.6 \Delta_{\mathrm{t}}$
Explanation:
(D) : The metal ion $\left(\mathrm{M}^{2+}\right)$ forms the complex with aqua ligands that means it contain the weak field ligand. It is a iron metal which has the following electronic configuration- $\mathrm{Fe}^{2 \mathrm{t}}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ If metal forms the tetrahedral complex then it contains the four unpaired electron. \(\text { CFSE }=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.4 \times \mathrm{N}_{\mathrm{t}_2}\right] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-0.6 \times 3+0.4 \times 3] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-1.8+1.2] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=-0.6 \Delta_{\mathrm{t}}\) Hence, the geometry and the crystal field stabilisation energy of the complex is tetrahedral and \(-0.6 \Delta_t\) respectively.
JEE Main 2020
COORDINATION COMPOUNDS
274289
The electronic spectrum of $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ shows a single broad peak with a maximum at 20,300 cm-1. The crystal field stabilization energy (CFSE) of the complex ion, in $\mathrm{kJ} \mathrm{mol}^{-1}$, is
1 145.5
2 242.5
3 83.7
4 97
Explanation:
(D) : Given: } \Delta_{\mathrm{o}}=20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=\text { ? }$ $\because \quad 1 \mathrm{Kj} / \mathrm{mol}=83.7 \mathrm{~cm}^{-1}$ ${\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Ti}$. $\therefore \quad x+6(0)=+3$ $x=+3$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{eg}_{\mathrm{g}}}\right] \Delta_{\mathrm{o}}$ $\mathrm{CFSE}=[-0.4 \times 1+0.6 \times 0] \Delta_{\mathrm{o}}$ Or $\mathrm{CFSE}=0.4 \Delta_{\mathrm{o}}$ (ignoring the negtive sign $)$ CFSE $=0.4 \times 20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=0.4 \times 20,300 \times \frac{1}{83.7} \mathrm{~kJ} / \mathrm{mol}$ $\mathrm{CFSE}=97 \mathrm{~kJ} \mathrm{~mol}^{-1}$
JEE Main 2020
COORDINATION COMPOUNDS
274290
The d-electron configuration of $\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$, respectively are
1 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
2 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
3 $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
4 $t_{2 g}^{4} e_{g}^{2}$ and $t_{2 g}^{4} e_{g}^{2}$
Explanation:
(B) : {$\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$} Let, $x$ be the oxidation state of Ru. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $\mathrm{x}$ be the oxidation state of Fe. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\because$ Pairing occure because 'Ru' belongs to the $4 \mathrm{~d} \quad \mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ (High spin) -series metal. $\mathrm{Ru}^{2+}[\mathrm{Kr}] 4 \mathrm{~d}^{6} 55^{\circ}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
JEE Main 2020
COORDINATION COMPOUNDS
274295
The complex that has highest crystal field splitting energy $(\Delta)$, is
(C) : It is clear that the CFSE of tetrahedral complex are always less than the octahedral complex. Thus, option (d) has less CFSE than option (a), (b) and (c) because it is a tetrahedral complex. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ forms the low spin complex because it contains the SFL. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ $3(+1)+\mathrm{x}+6(-1)=0$ $\mathrm{x}=+3$ $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ $\mathrm{d}^{6}-\mathrm{C}_{\mathrm{g}}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 g}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{\text {o }}$ $\mathrm{CFSE}=[-0.4 \times 6+0.6 \times 0] \Delta_{\text {。 }}$ $\mathrm{CFSE}=-2.4 \Delta_{\mathrm{o}}$ Hence, $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ has highest crystal field splitting energy.
JEE Main 2019
COORDINATION COMPOUNDS
274296
The crystal field stabilization energy (CFSE) of $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$, respectively, are
1 $-0.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
2 $-0.4 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
3 $-2.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
4 $-0.6 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
Explanation:
(B) : $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{NiCl}_{4}\right]$ are octahedral and tetrahedral complex respectively. $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $x$ be the oxidation state of Fe. $x+6(0)+2(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{0}$ CFSE $=[-0.4 \times 4+0.6 \times 2] \Delta_{0}$ $\mathrm{CFSE}=-0.4 \Delta_{0}$ and, for $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$ $2(+1)+x+4(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}$ $\mathrm{d}^{8}-\uparrow_{\uparrow \downarrow} \uparrow \uparrow_{\mathrm{t}_{2}}$ CFSE $=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.6 \times \mathrm{N}_{\mathrm{t}_{2}}\right] \Delta_{\mathrm{t}}$ CFSE $=[-0.6 \times 4+0.6 \times 4] \Delta_{\mathrm{t}}$ $\mathrm{CFSE}=-0.8 \Delta_{\mathrm{t}}$
274288
Consider that a $\mathrm{d}^{6}$ metal ion $\left[\mathrm{M}^{2+}\right]$ forms a complex with aqua ligands and the spin only magnetic moment of the complex is 4.90 BM. The geometry and the crystal field stabilisation energy of the complex is
1 tetrahedral and $-1.6 \Delta_{\mathrm{t}}+1 \mathrm{P}$
2 octahedral and $-2.4 \Delta_{0}+2 \mathrm{P}$
3 octahedral and $-1.6 \Delta_{0}$
4 tetrahedral and $-0.6 \Delta_{\mathrm{t}}$
Explanation:
(D) : The metal ion $\left(\mathrm{M}^{2+}\right)$ forms the complex with aqua ligands that means it contain the weak field ligand. It is a iron metal which has the following electronic configuration- $\mathrm{Fe}^{2 \mathrm{t}}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ If metal forms the tetrahedral complex then it contains the four unpaired electron. \(\text { CFSE }=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.4 \times \mathrm{N}_{\mathrm{t}_2}\right] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-0.6 \times 3+0.4 \times 3] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=[-1.8+1.2] \Delta_{\mathrm{t}}\) \(\mathrm{CFSE}=-0.6 \Delta_{\mathrm{t}}\) Hence, the geometry and the crystal field stabilisation energy of the complex is tetrahedral and \(-0.6 \Delta_t\) respectively.
JEE Main 2020
COORDINATION COMPOUNDS
274289
The electronic spectrum of $\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}$ shows a single broad peak with a maximum at 20,300 cm-1. The crystal field stabilization energy (CFSE) of the complex ion, in $\mathrm{kJ} \mathrm{mol}^{-1}$, is
1 145.5
2 242.5
3 83.7
4 97
Explanation:
(D) : Given: } \Delta_{\mathrm{o}}=20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=\text { ? }$ $\because \quad 1 \mathrm{Kj} / \mathrm{mol}=83.7 \mathrm{~cm}^{-1}$ ${\left[\mathrm{Ti}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}}$ Let, $\mathrm{x}$ be the oxidation state of $\mathrm{Ti}$. $\therefore \quad x+6(0)=+3$ $x=+3$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{eg}_{\mathrm{g}}}\right] \Delta_{\mathrm{o}}$ $\mathrm{CFSE}=[-0.4 \times 1+0.6 \times 0] \Delta_{\mathrm{o}}$ Or $\mathrm{CFSE}=0.4 \Delta_{\mathrm{o}}$ (ignoring the negtive sign $)$ CFSE $=0.4 \times 20,300 \mathrm{~cm}^{-1}$ $\mathrm{CFSE}=0.4 \times 20,300 \times \frac{1}{83.7} \mathrm{~kJ} / \mathrm{mol}$ $\mathrm{CFSE}=97 \mathrm{~kJ} \mathrm{~mol}^{-1}$
JEE Main 2020
COORDINATION COMPOUNDS
274290
The d-electron configuration of $\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$ and $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$, respectively are
1 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
2 $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ and $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
3 $\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$ and $\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$
4 $t_{2 g}^{4} e_{g}^{2}$ and $t_{2 g}^{4} e_{g}^{2}$
Explanation:
(B) : {$\left[\mathrm{Ru}(\mathrm{en})_{3}\right] \mathrm{Cl}_{2}$} Let, $x$ be the oxidation state of Ru. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $\mathrm{x}$ be the oxidation state of Fe. $\therefore \mathrm{x}+3(0)+2(-1)=0$ $\mathrm{x}=+2$ $\because$ Pairing occure because 'Ru' belongs to the $4 \mathrm{~d} \quad \mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ (High spin) -series metal. $\mathrm{Ru}^{2+}[\mathrm{Kr}] 4 \mathrm{~d}^{6} 55^{\circ}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{6} \mathrm{e}_{\mathrm{g}}^{0}$ d-electron configuration $=\mathrm{t}_{2 \mathrm{~g}}^{4} \mathrm{e}_{\mathrm{g}}^{2}$
JEE Main 2020
COORDINATION COMPOUNDS
274295
The complex that has highest crystal field splitting energy $(\Delta)$, is
(C) : It is clear that the CFSE of tetrahedral complex are always less than the octahedral complex. Thus, option (d) has less CFSE than option (a), (b) and (c) because it is a tetrahedral complex. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ forms the low spin complex because it contains the SFL. $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ $3(+1)+\mathrm{x}+6(-1)=0$ $\mathrm{x}=+3$ $\mathrm{Co}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ $\mathrm{d}^{6}-\mathrm{C}_{\mathrm{g}}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 g}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{\text {o }}$ $\mathrm{CFSE}=[-0.4 \times 6+0.6 \times 0] \Delta_{\text {。 }}$ $\mathrm{CFSE}=-2.4 \Delta_{\mathrm{o}}$ Hence, $\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]$ has highest crystal field splitting energy.
JEE Main 2019
COORDINATION COMPOUNDS
274296
The crystal field stabilization energy (CFSE) of $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$, respectively, are
1 $-0.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
2 $-0.4 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
3 $-2.4 \Delta_{0} \text { and }-1.2 \Delta_{\mathrm{t}}$
4 $-0.6 \Delta_{0} \text { and }-0.8 \Delta_{\mathrm{t}}$
Explanation:
(B) : $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ and $\mathrm{K}_{2}\left[\mathrm{NiCl}_{4}\right]$ are octahedral and tetrahedral complex respectively. $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Cl}_{2}$ Let, $x$ be the oxidation state of Fe. $x+6(0)+2(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Fe}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{6}$ CFSE $=\left[-0.4 \times \mathrm{N}_{\mathrm{t}_{2 \mathrm{~g}}}+0.6 \times \mathrm{N}_{\mathrm{e}_{\mathrm{g}}}\right] \Delta_{0}$ CFSE $=[-0.4 \times 4+0.6 \times 2] \Delta_{0}$ $\mathrm{CFSE}=-0.4 \Delta_{0}$ and, for $\mathrm{K}_{2}\left[\mathrm{Ni}(\mathrm{Cl})_{4}\right]$ $2(+1)+x+4(-1)=0$ $\mathrm{x}=+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 \mathrm{~d}^{8}$ $\mathrm{d}^{8}-\uparrow_{\uparrow \downarrow} \uparrow \uparrow_{\mathrm{t}_{2}}$ CFSE $=\left[-0.6 \times \mathrm{N}_{\mathrm{e}}+0.6 \times \mathrm{N}_{\mathrm{t}_{2}}\right] \Delta_{\mathrm{t}}$ CFSE $=[-0.6 \times 4+0.6 \times 4] \Delta_{\mathrm{t}}$ $\mathrm{CFSE}=-0.8 \Delta_{\mathrm{t}}$
